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Find integer solutions for m - n(log3(2)) = 10(log9(6)).

From the properties of logarithms (logba = logca / logcb), 10log96 can be rewritten as 10(log36 / log39). Since log39 = 2, 10log96 = 5log36. We then bring both log terms to the same side of the equation: m = 5log36 + nlog32. Again, from log properties (a(logcb) = logcba), this can be rewritten as m = log365 + log32n. Since the logarithms have the same base, we can combine them using another log property (logab + logac = logabc). This yields m = log3652n. We can factor out the 25 from the 65 to obtain m = log335252n. Combining the 2s: m = log33525 + n. We then raise 3 to the power of both sides to get an equation without logarithms: 3m = 3525 + n. We can write an invisible 20 term on the left side without changing the equation, giving 3m20 = 3525 + n. Since m and n are integers, m = 5 and 5 + n = 0, meaning n = -5. So the solution is m = 5, n = -5.

Answered by Tristan K. Maths tutor

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